finite projective planes have q2+q+1 points and q2+q+1 lines

Given a finite projective plane that contains a quadrangle OXYZ (i.e. no three of these four points are on one line). To prove: the plane has q2+q+1 points and q2+q+1 lines for some integer q, and there are q+1 points on each line and q+1 lines through each point.

Let x and y be the lines OX and OY, which must exist by the axioms. By the assumptionPlanetmathPlanetmath OXYZ is a quadrangle these lines are distinct and Z is not on them. Let there be p points Xi on x other than O, for each of them one line ZXi exists, and is distinct (one lines cannot pass through two Xi unless it is x but that’s not a line through Z). Conversely every line through Z must intersect x in a unique point (two lines intersecting in Z cannot intersect at another point, and Z is not a point on x). So there are p+1 lines through Z (OZ is one of them). By the same reasoning, using y, there are q+1 lines through Z so p=q. We also found q+1 points (including O) on y and the same number on x. Intersecting the q+1 lines through Z with XY (on which Z does not lie, the quadrangle again) reveals at least q+1 distinct points there and at most q+1 because for each point there there is a line through it and Z.

The lines not through O intersect x in one of the q points Xi and y in one of the q points Yj. There are q2 possibilities and each of them is a distinct line, because there is only one line through a given Xi and Yj. The lines that do pass through O intersect XY in one of the q+1 points there, again one line for each such point and vice versa. That’s q+1 lines through O and q2 not through O, q2+q+1 in all.

There are q+1 lines through X (to each of the points of y) and q+1 lines through Y (to each of the points of x). Intersect the q lines through X other than XY with the q lines through Y other than XY, these q2 intersections are all distinct because for any P there’s only one line PX and one line PY. Note we did not use the line XY. Conversely for any P not on XY there must be some PX and some PY, so there are exactly q2 points not on XY. Add the q+1 points on XY for a total of q2+q+1.

The constructions above already showed q+1 lines through some points (X, Y and Z), by the same games as before that implies for each of them q+1 points on every line not through that point. We also saw q+1 points on some lines (x, y, XY) which implies for each of them q+1 lines through every point not on that line. Such reasoning covers q2 items on first application and rapidly mops up stragglers on repeated application.

Some form of this proof is standard math lore; this version was half remembered and half reconstructed.

Title finite projective planes have q2+q+1 points and q2+q+1 lines
Canonical name FiniteProjectivePlanesHaveQ2q1PointsAndQ2q1Lines
Date of creation 2013-03-22 15:11:18
Last modified on 2013-03-22 15:11:18
Owner marijke (8873)
Last modified by marijke (8873)
Numerical id 4
Author marijke (8873)
Entry type Proof
Classification msc 51A35
Classification msc 05B25
Classification msc 51E15